# What is the remainder when a polynomial $g(x^{12})$ is divided by $g(x)$?

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## Solution 1

HINT :

$$g(x)=x^5+x^4+x^3+x^2+x+1=\frac{x^6-1}{x-1}$$ for $x\not=1$ and \begin{align}g(x^{12})&=x^{60}+x^{48}+x^{36}+x^{24}+x^{12}+1\\&=(x^{60}-1)+(x^{48}-1)+(x^{36}-1)+(x^{24}-1)+(x^{12}-1)+6\end{align}

## Solution 2

We have $$g(x^{12}) = (x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1 ) = (x^5 + x^4 + x^3 + x^2 + x^1 + 1) \cdot q(x) + r(x)$$ where $r(x)$ is a polynomial with degree no greater than four.

To find $r(x)$, we can plug in the roots of $g(x)$, the non-one sixth roots of unity, into the equation. Let these roots be $\omega, \omega^2, \omega^3, \omega^4, \omega^5$. Note that for each of these, $x^{60} = x^{48} = x^{36} = x^{24} = x^{12} = 1$. Therefore, for any $i$, $r(x) - 6 = 0$ for all values of $\omega^i$.

Therefore, $r(x) - 6$ is a with five roots and degree no greater than four, and it follows that we must have $r(x) = 6$.

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### Karthik Macherla

Updated on October 12, 2022

### Comments

• Karthik Macherla 7 months

Let $g(x) = x^5 +x^4 +x^3+x^2+x+1$.

What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$?

• André Nicolas almost 8 years
Note that if $t\ne 1$ then $g(t)=\frac{t^6-1}{t-1}$.